Question

If the angles of a triangle are ${30}^o$ and ${45}^o$ and the included side is $(\sqrt{3}+1)$ , then area of $\Delta$ is :

• A. $\frac{1}{\sqrt{3}-1}$
• B. $\sqrt{3}+1$
• C. $\frac{1}{\sqrt{3}+1}$
• D. None

Answer 1

The correct option is A $\frac{1}{\sqrt{3}-1}$

Draw $AD\perp BC,BC=\sqrt{3}+1$

Let $BD+x$ then $CD=\sqrt{3}+1-x$

In $△ABD,\tan {30}^o=\frac{AD}{BD}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AD}{x}\Rightarrow x=\sqrt{3}AD⟶\left(1\right)$

Now, in $△ADC,\tan {45}^o=\frac{AD}{CD}$

$1=\frac{AD}{\sqrt{3}+1-x}$

$\Rightarrow \sqrt{3}+1-x=AD$

From $\left(1\right),$ put the value of $x$, we get

$\Rightarrow \sqrt{3}+1-\sqrt{3}AD=AD$

$\Rightarrow (\sqrt{3}+1)\Rightarrow AD=1$

ar $△ABC=\frac{1}{2}\times BC\times AD=\frac{1}{2}\times \sqrt{3}+1\times 1$

$=\frac{\sqrt{3}+1}{2}=\frac{(\sqrt{3}+1)(\sqrt{3}-1)}{2(\sqrt{3}-1)}$

$=\frac{3-1}{2(\sqrt{3}-1)}=\frac{1}{\sqrt{3}-1}$